A solar power park of 12 km² is planned in Rajasthan. The average solar insolation is 6.5 kWh m⁻² day⁻¹, the photovoltaic panels have an efficiency of 20%, and the plant is expected to operate 310 days a year. Calculate the total electricity that can be generated in a year in gigawatt‑hours.
Answer
4836 GWh
The correct answer is 4836 GWh — Convert to GWh: 4.836 × 10⁹ kWh ÷ 10⁶ = 4.836 × 10³ GWh ≈ 4836 GWh.
Step-by-step solution
- Convert area: 12 km² = 12 × 10⁶ m².
- Daily incident energy per m² = 6.5 kWh.
- Apply panel efficiency: 6.5 × 0.20 = 1.3 kWh/m²/day.
- Total daily energy = 1.3 × 12 × 10⁶ = 15.6 × 10⁶ kWh.
- Annual energy = 15.6 × 10⁶ × 310 = 4.836 × 10⁹ kWh.
- Convert to GWh: 4.836 × 10⁹ kWh ÷ 10⁶ = 4.836 × 10³ GWh ≈ 4836 GWh.
Common mistake: Forgetting to apply panel efficiency or operation days
Concept tested: Solar energy generation calculations
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